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A Simple Derivation ofHadamard’s Variational Formula
Nuha Loling OTHMAN, 1 Takashi SUZUKI (鈴木 貴 ), 2 Takuya TSUCHIYA
(土屋 卓也), 3
ここでは、 $\mathbb{R}^{n}$ 内の有界領域上での Poisson 問題と、 その問題の Green
関数を考える。約 100年前、Hadamard は、領域の境界が摂動を受けた際に Green
関数がどのような影響を受けるかという問題を考え、領域の摂動に対する Green 関数の第一変分を求めた。 それは、 現在
Hadamard の変分公式と呼ばれている。 この論文では、 Hadamard の変分公式の別証明を与える。
さらに、領域の摂動に対する Green 関数の第二変分も計算することができた。我々の公式は、
Garabedian-Schifferの公式の拡張になっている。
1 Introduction
Let $\mathbb{R}^{n}$ be n-dimensional Euclidean space $(n\geq
2)$ and $\Omega\subset \mathbb{R}^{n}$ be a bounded domain. For
agiven function $f$ , we consider Poisson’s equation
$-\triangle u=f$ on $\Omega$ , $u=0$ on $\partial\Omega$ .
The Green function $G(x, y)$ is a function which provides the
solution $u$ of the Poisson equationby
$u(x)= \int_{\Omega}G(x, y)f(y)dy$ .
If the domain $\Omega$ is modified, then the Green function
$G(x, y)$ would vary. Hadamard consideredhow $G(x, y)$ would vary
and computed the first variation $\delta G(x, y)$ with respect to
domain per-turbation [3]. His result is now called Hadamard’s
variational formula. Hadamard showedhis formula under the
assumption that $\partial\Omega$ and the perturbation are analytic.
Later, Garabedianand Schiffer gave a simpler and more rigorous
proof of Hadamard‘s variational formula underthe assumption that
$\partial\Omega$ and the perturbation are of $C^{2}$ class (see
[1]). Further, they obtainedHadamard’s second variational formula
[2], [4]. The main aim of this paper is to reconsiderHadamard’s
variational formula. In particular, we develop a methodology which
provides us amuch clearer understanding of Hadamard’s variational
formula. As a result, we obtain a verysimple proof of Hadamard’s
variational formula (see Section 3.1). We also obtain
Hadamard’ssecond variational formula which is an extension of
Grabedian-Schiffer’s formula (Theorem 3.3).
Here, we briefly summarize the notation which we use in this
paper. We denote the Euclideaninner product by $x\cdot y$ or $(x,
y)_{\mathbb{R}^{n}}$ for $x,$ $y\in \mathbb{R}^{n}$ . When we do
not specify, all vectors in $\mathbb{R}^{n}$ areregarded as column
vectors. Transposing of vectors and matrices are denoted by
$(\cdot)^{T}$ . Let $f(x)$be a smooth function defined in a domain
of $\mathbb{R}^{n}$ . The gradient of $f$ is denoted by
$\nabla f(x):=(\frac{\partial f}{\partial x_{1}}(x),$
$\cdots,$$\frac{\partial f}{\partial x_{n}}(x))$ .
lGraduate School of Engineering Science, Osaka University,
othmanQsigmath. es. osaka-u. ac. jp2Graduate School of Engineering
Science, Osaka University, suzukiQsigmath. es. osaka-u. ac.
jp3Graduate School of Science and Engineering, Ehime University,
tsuchiyaQmath. sci. ehime-u. ac. jp
数理解析研究所講究録第 1733巻 2011年 127-141 127
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When we need specify the variable of a gradient, we denote such
as $\nabla_{x}f(x),$ $\nabla_{x}*f(x^{*})$ . Weregard gradients of
functions as row vectors. Hence, for a vector field $F(x),$ $\nabla
F(x)$ is the Jacobimatrix $DF(x)$ . Let $\Omega\subset
\mathbb{R}^{n}$ be a domain in $\mathbb{R}^{n}$ . We denote by
$L^{2}(\Omega),$ $H^{1}(\Omega),$ $H^{s}(\partial\Omega)$ the
usualLebesgue and Sobolev spaces. The inner product of
$L^{2}(\Omega)$ is denoted by
$(u, v)_{\Omega}$ $:= \int_{\Omega}$ uvdx, $u,$ $v\in
L^{2}(\Omega)$
On a point $x\in\partial\Omega$ , we denote the unit outer
normal vector of $\partial\Omega$ by $\nu=\nu(x)$ . For asubset
$\Gamma\subset\partial\Omega$ , we denote the duality pair of
$H^{-1/2}(\Gamma)$ and $H^{1/2}(\Gamma)$ by $\langle\varphi,$
$v\rangle_{\Gamma},$ $\varphi\in H^{-1/2}(\Gamma)$ ,$v\in
H^{1/2}(\Gamma)$ .
2 Basic Definitions
Let $\Omega\subset \mathbb{R}^{n}$ be a bounded Lipschitz domain
and $\tilde{\Omega}$ be a sufficiently larger domain which
satisfies$\overline{\Omega}\subset$ int $\tilde{\Omega}$ . For a
parameter $t\geq 0$ , we define transformation $\mathcal{T}_{t}$ :
$\Omegaarrow \mathcal{T}_{t}(\Omega)\subset \mathbb{R}^{n}$ of
$\Omega$with respect $t$ in the following way. Let a $C^{0,1}$
-class vector field $S(x)$ be given. We supposethat suppS
$\subset\tilde{\Omega}$ . Then, a transformation
$\mathcal{T}_{t}(x)$ on fi is defined as a solution of the
ordinarydifferential equation
(2.1) $\frac{d}{dt}\mathcal{T}_{t}(x)=S(\mathcal{T}_{t}(x))$ ,
$\mathcal{T}_{0}(x)=x$ .
That is, for each $x\in\tilde{\Omega},$ $\mathcal{T}_{t}(x)$ is
the integral curve generated by (2.1). This $\mathcal{T}_{t}(x)$
satisfies thefollowing properties:
$\bullet$ For any $x\in\Omega,$ $T_{0}(x)=x$ .. For a
sufficiently small $t,$ $\Omega_{t}$
$:=\mathcal{T}_{t}(\Omega)\subset\tilde{\Omega}$ .$\bullet$
$\mathcal{T}_{t}$ is a diffeomorphism for a sufficiently small
$t\geq 0$ .
$\bullet$ $\mathcal{T}_{t}$ is smooth with respect to $t$ .
From the definition (2.1) we have $S(x)=\frac{\partial}{\partial
t}\mathcal{T}_{t}(x)|_{t=0}$ . Moreover, we define
$T(x):=\frac{\partial^{2}}{\partial
t^{2}}\mathcal{T}_{t}(x)|_{t=0}$ .
Then, the transformation has the Taylor expansion
$\mathcal{T}_{t}(x)=x+tS(x)+\frac{1}{2}t^{2}T(x)+o(t^{2})$
with respect to $t$ . Here, $o(t^{2})$ denote a quantity which
would be expressed by $t^{2}\omega(x, t)$ , where$\omega(x, t)$ is
a function which converges uniformly (with respect to x) to $0$ as
$tarrow+O$ . In the sequel,notations such as $o(t),$ $o(t^{2})$ are
understood in this way. Let $DS(x)$ be the Jacobi matrix of S.From
(2.1), we have
$\frac{d^{2}}{dt^{2}}\mathcal{T}_{t}(x)=\frac{d}{dt}S(\mathcal{T}_{t}(x))=DS(\mathcal{T}_{t}(x))\frac{d}{dt}\mathcal{T}_{t}(x)=DS(\mathcal{T}_{t}(x))S(\mathcal{T}_{t}(x))$
,
which implies
(2.2) $T(x)=(DS(x))S(x)$ .
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Let a function $\varphi$ be defined on $\tilde{\Omega}$ and
$\varphi\in H^{2}(\tilde{\Omega})$ . Suppose that a function
$u=u(x, t)\in$$H^{1}(\Omega_{t})$ is a solution of the boundary
value problem
(2.3) $\{\begin{array}{ll}\triangle v(\cdot, t)=0 in
\Omega_{t},u(\cdot, t)=\varphi on
\partial\Omega_{t}.\end{array}$
Here, $\triangle$ $:=\partial^{2}/\partial
x_{1}^{2}+\cdots+\partial^{2}/\partial x_{n}^{2}$ is the usual
Laplacian with respect to $x=(x_{1}, \cdots, x_{n})^{T}$ . In
thissection, we investigate differentiations of quantities which
depend on $u(x, t)$ . Such variations ofquantities with respect to
domain perturbation are called Hadamard $s$ variation. To
computeHadamard’s variation it is important to know Lagrangian
derivative 4 $\dot{u}_{\mathcal{L}},$ $ii_{\mathcal{L}}$ and
Eulerianderivative 5 $\dot{u}\mathcal{E},\ddot{u}\mathcal{E}$ ,
defined by, for $x\in\Omega$ ,
$\dot{u}_{\mathcal{L}}(x):=\frac{d}{dt}(u(\mathcal{T}_{t}(x),
t))|_{t=0}$
$\ddot{u}_{\mathcal{L}}(x):=\frac{d^{2}}{dt^{2}}(u(\mathcal{T}_{t}(x),
t))|_{t=0}$ ,
$\dot{u}_{\mathcal{E}}(x):=\frac{\partial}{\partial t}u(x,
t)|_{t=0}$ ,
$\ddot{u}_{\mathcal{E}}(x):=\frac{\partial^{2}}{\partial t^{2}}u(x,
t)|_{t=0}$ .
For a function $f(x, t)$ , let
$\mathcal{H}_{x}f=\mathcal{H}_{x}f(x,
t):=(\frac{\partial^{2}f(x,t)}{\partial x_{i}\partial
x_{j}})_{i,j=1,\cdots,n}$
be the Hesse matrix. We use the same notation $\mathcal{H}_{x}f$
for the second order tensor $\mathcal{H}_{x}f$ :
$\mathbb{R}^{n}\cross \mathbb{R}^{n}arrow$$\mathbb{R}$ defined by
$\mathcal{H}_{x}f(X, Y)$ $:=((\mathcal{H}_{x}f)X,
Y)_{\mathbb{R}^{n}}$ for $X,$ $Y\in \mathbb{R}^{n}$ . In
particular, in the case of $X=Y$ ,we denote as $\mathcal{H}_{x}f(X,
X)=\mathcal{H}_{x}f\cdot(X)^{2}$ . A straightforward computation
yields
(2.4) $\frac{d}{dt}(u(\mathcal{T}_{t}(x),
t))=\frac{\partial}{\partial t}u(\mathcal{T}_{t}(x), t)+\nabla
u(\mathcal{T}_{t}(x), t)\cdot(\frac{\partial}{\partial
t}\mathcal{T}_{t}(x))$ ,
(2.5) $\frac{d^{2}}{dt^{2}}(u(\mathcal{T}_{t}(x),
t))=\frac{\partial^{2}}{\partial t^{2}}u(\mathcal{T}_{t}(x),
t)+2\nabla(\frac{\partial}{\partial t}u(\mathcal{T}_{t}(x),
t))\cdot(\frac{\partial}{\partial t}\mathcal{T}_{t}(x))$
$+ \nabla u(\mathcal{T}_{t}(x),
t)\cdot(\frac{\partial^{2}}{\partial
t^{2}}\mathcal{T}_{t}(x))+\mathcal{H}_{x}u(\mathcal{T}_{t}(x),
t)\cdot(\frac{\partial}{\partial t}\mathcal{T}_{t}(x))^{2}$ .
2.1 Eulerian Derivatives
$\dot{u}_{\mathcal{E}},\ddot{u}_{\mathcal{E}}$
In this subsection, we check properties which Eulerian
derivatives $\dot{u}_{\mathcal{E}},\ddot{u}_{\mathcal{E}}$ should
satisfy. At aninner point $x\in\Omega$ we have $\triangle u(\cdot,
t)=0$ for any $t$ . Hence,
$\triangle\dot{u}_{\mathcal{E}}=0$ ,
$\triangle\ddot{u}_{\mathcal{E}}=0$ in $\Omega$ .
On the boundary $\partial\Omega$ we have $u(\mathcal{T}_{t}(x),
t)=\varphi(\mathcal{T}_{t}(x))$ . Differentiating the both side and
letting$tarrow+0$ , we see $\dot{u}_{\mathcal{E}}+$ S. $\nabla u=$
S. $\nabla\varphi$ . Therefore, we find that the Eulerian
derivative $\dot{u}s$ is asolution of the following boundary value
problem:
(2.6) $\triangle\dot{u}\mathcal{E}=0$ in $\Omega$ ,
$\dot{u}\mathcal{E}=S$ . $(\nabla\varphi-\nabla u)$ on
$\partial\Omega$ .
In the same manner, we conclude that $\ddot{u}_{\mathcal{E}}$ is
a solution of the boundary value
problem$\triangle\ddot{u}_{\mathcal{E}}=0$ in $\Omega$ ,
(2.7)$\ddot{u}_{\mathcal{E}}=-2S$ .
$\nabla\dot{u}_{\mathcal{E}}+T$ . $(\nabla\varphi-\nabla
u)+(\mathcal{H}_{x}\varphi-\mathcal{H}_{x}u)\cdot(S)^{2}$ on
$\partial\Omega$ .
4It is also called material derivative or covariant
demvative.5This ib a usual partial derivative with respect to $t$
which is also called shape derivative.
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2.2 Lagrangian Derivatives
$\dot{u}_{\mathcal{L}},\ddot{u}_{\mathcal{L}}$
In this subsection, we check properties which Lagrangian
derivatives $\dot{u}_{\mathcal{L}},\ddot{u}_{C}$ should satisfy.
Here,variable on $\Omega_{t}$ is denoted as
$x^{*}=\mathcal{T}_{t}(x)$ . A function $f(x^{*})$ defined on
$\Omega_{t}$ is pulled back by $\mathcal{T}_{t}$ toa function
$f(x)$ on $\Omega$ as
$f(x):=f(T_{t}(x))$ .
Note that we have
$\nabla_{x’}f=(\frac{\partial f}{\partial x_{1}},$$\cdots,$
$\frac{\partial f}{\partial
x_{n}})(\begin{array}{lll}\frac{\partial x\iota}{\partial
x_{1}^{*}} \frac{\partial x1}{\partial x_{7l}}| .
|\frac{\partial}{\partial}xAx_{1}^{*} \cdots
\frac{\partial}{\partial}x_{A}x_{n}^{*}\end{array})=(
\nabla_{x}f)(D\mathcal{T}_{t}^{-1})$ ,
where $DT_{t}^{-1}$ is the Jacobi matrix of
$\mathcal{T}_{t}^{-1}$ .The weak form of the boundary value problem
(2.3) is
(2.8) $\{\begin{array}{l}(\nabla u(\cdot, t),
\nabla\tilde{v})_{\Omega_{t}}=0, \forall\tilde{v}\in
H_{0}^{1}(\Omega_{t}),u(\cdot, t)=\varphi on
\partial\Omega_{t}.\end{array}$
Using the transformation $\mathcal{T}_{t}$ , we pull back the
problem (2.8) to a problem defined on $\Omega$ . Notethat
$\overline{\tau)}\in H_{0}^{1}(\Omega_{t})\Leftrightarrow
v:=\tilde{v}\circ \mathcal{T}_{t}\in H_{0}^{1}(\Omega)$.
Then, setting $u_{t}(x);=u(\mathcal{T}_{t}(x), t)$ , we see
that
$( \nabla u(\cdot, t),
\nabla\tilde{v})_{\Omega_{f}}=\int_{\Omega}(\det DT_{t})(\nabla
u_{t}(D\mathcal{T}_{t}^{-1}0\mathcal{T}_{t})(D\mathcal{T}_{t}^{-1}0\mathcal{T}_{t})^{T})\cdot\nabla
vdx$
$=(A(t)\nabla u_{t}, \nabla\iota))_{\Omega}$ , $\forall v\in
H_{0}^{1}(\Omega)$ ,
where$A(t)$ $:=(\det
D\mathcal{T}_{t})(D\mathcal{T}_{t}^{-1}0\mathcal{T}_{t})(D\mathcal{T}_{t}^{-1}0\mathcal{T}_{t})^{T}$
.
That is, the boundary value problem (2.8) on $\Omega_{t}$ is
pulled back to the boundary value problem
(2.9) $\{\begin{array}{l}(A(t)\nabla u_{t}, \nabla
v)_{\Omega}=0, \forall v\in H_{0}^{1}(\Omega),u_{t}=\varphi oT_{t}
on \partial\Omega\end{array}$
on $\Omega$ . If $u(x, t)$ is a solution of (2.8), then
$u_{t}(x)=u(\mathcal{T}_{t}(x), t)$ is a solution of (2.9) and vice
versa.We set
(2.10) $\mathcal{A}’:=\frac{d}{dt}A(t)|_{t=0}$ ,
$\mathcal{A}’’:=\frac{d^{2}}{dt^{2}}A(t)|_{t=0}$ .
Suppose that $\varphi 0\mathcal{T}_{t}$ has the following Taylor
expansion:
$\varphi\circ
\mathcal{T}_{t}=\varphi+t\dot{\varphi}+\frac{1}{2}t^{2}\ddot{\varphi}+o(t^{2})$
.
From the definition we find
$\dot{\varphi}=S\cdot\nabla\varphi$ .
$\ddot{\varphi}=T\cdot\nabla_{\forall^{\neg}}’+\mathcal{H}_{x}\varphi\cdot(S)^{2}$
.
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Let $u$ be a solution of (2.3). Then, we ($(differentiate$“
(2.9) and obtain the equation
(2.11) $\{\begin{array}{l}(\nabla\dot{u}c, \nabla
v)_{\Omega}=-(\mathcal{A}’\nabla u, \nabla c))_{\Omega},\forall
v\in H_{0}^{1}(\Omega),\dot{u}_{\mathcal{L}}=(i on
\partial\Omega.\end{array}$
One more ($(differentiation$“ yields the equation
(2.12)$(\nabla\ddot{u}_{\mathcal{L}}, \nabla
v)_{\Omega}=-2(\mathcal{A}^{f}\nabla\dot{u}_{\mathcal{L}}, \nabla
v)_{\Omega}-(\mathcal{A}’’\nabla u, \nabla v)_{\Omega}$, $\forall
v\in H_{0}^{1}(\Omega)$ ,
$\ddot{u}c=\ddot{\varphi}$ on $\partial\Omega$ .
For solutions of these equations, we have the following
lemma.
Lemma 2.1 Suppose that $u,$
$u_{t},\dot{u}_{\mathcal{L}},\ddot{u}_{\mathcal{L}}\in
H^{1}(\Omega)$ are solutions of the equations (2.3), (2.9),(2.11),
(2.12), respectively. Then, $u_{t}$ has a Taylor expansion
$u_{t}=u+t
\dot{u}_{\mathcal{L}}+\frac{1}{2}t^{2}\ddot{u}c+o(t^{2})$
in$H^{1}(\Omega)$ . That is, the following is valid;
$\lim_{tarrow
0+}\frac{\Vert\chi_{t}||_{H^{1}(\Omega)}}{t^{2}}=0$,
$\chi_{t}:=v_{t}-(u+t\dot{u}_{\mathcal{L}}+\frac{1}{2}t^{2}ii_{\mathcal{L}})$
.
Proof Since $S\in W^{1,\infty}(\tilde{\Omega};\mathbb{R}^{n})$
and $T\in L^{\infty}(\tilde{\Omega};\mathbb{R}^{n})$ , we see
$A(t)\in L^{\infty}(\Omega;\mathbb{R}^{n^{2}})$ and
(2.13) $\lim_{tarrow
0+}\frac{\Vert\alpha_{t}||_{L^{\infty}}}{t^{2}}=0$ ,
$\alpha_{t}:=A(t)-(1+t\mathcal{A}’+\frac{1}{2}t^{2}\mathcal{A}’’)$
.
Define $z_{t},\dot{z},\ddot{z}$ as solutions of the following
boundary value problems:
$(\nabla z_{t}, \nabla v)_{\Omega}=0$ , $\forall v\in
H_{0}^{1}(\Omega)$ , $z_{t}=\varphi 0\mathcal{T}_{t}$ on
$\partial\Omega$ ,$(\nabla\dot{z}, \nabla v)_{\Omega}=0$ , $\forall
v\in H_{0}^{1}(\Omega)$ , $\dot{z}=\dot{\varphi}$ on
$\partial\Omega$ ,$(\nabla\ddot{z}, \nabla v)_{\Omega}=0$ ,
$\forall v\in H_{0}^{1}(\Omega)$ , $\sim\ddot{\vee}=\ddot{\varphi}$
on $\partial\Omega$ .
Letting$\eta_{t}:=z_{t}-(u+t\dot{z}+\frac{1}{2}t_{\sim}^{2_{\vee}})$
, $\psi_{t}:=\varphi\circ
\mathcal{T}_{t}-(\varphi+t\dot{\varphi}+\frac{1}{2}t^{2}\ddot{\varphi})$
,
we notice $\eta_{t}-\psi_{t}\in H_{0}^{1}(\Omega)$ . Since
$0=(\nabla\eta_{t}, \nabla v)_{\Omega}$ for any $v\in
H_{0}^{1}(\Omega)$ , we set $v$ $:=77t-\psi_{t}$ andobtain
$\Vert\nabla\eta_{t}\Vert_{L^{2}(\Omega)}^{2}=(\nabla\eta_{t},
\nabla\eta_{t})_{\Omega}=(\nabla\eta_{t},
\nabla?l_{t}))_{\Omega}\leq\Vert\nabla_{7}h\Vert_{L^{2}(\Omega)}\Vert\nabla?i_{t}\Vert_{L^{2}(\Omega)}$
,
$\lim_{tarrow
0+}\frac{\Vert\nabla\eta_{t}||_{L^{2}(\Omega)}}{t^{2}}\leq\lim_{tarrow
0+}\frac{\Vert\nabla\psi_{t}||_{L^{2}(\Omega)}}{t^{2}}=0$ .
Similarly, set$\beta_{t}$ $:=$ 娩一 $z_{t}-$ $(t$ $(\dot{u}c$ 一ゑ
$)$ $+ \frac{1}{2}t^{2}(\ddot{u}_{C}-\sim\vee))\in
H_{0}^{1}(\Omega)$
Then, from (2.11), (2.12), we find that for any $v\in
H_{0}^{1}(\Omega)$ ,
$(A(t)\nabla\beta_{t}, \nabla v)_{\Omega}=((l-A(t))\nabla z_{t},
\nabla v)_{\Omega}$
$-t( A(t)\nabla(\dot{u}c-\dot{z}), \nabla
v)_{\Omega}-\frac{1}{2}t^{2}(A(t)\nabla(\ddot{u}_{\mathcal{L}}-$ を
$), \nabla v)_{\Omega}$
$=((I- A(t))\nabla(\eta_{t}+\frac{1}{2}t^{2}\ddot{u}c), \nabla
v)_{\Omega}+t((I+tA’-A(t))\nabla\dot{u}c, \nabla v)_{\Omega}$
$+$ $((I+t \mathcal{A}’+\frac{1}{2}t^{2}\mathcal{A}’’ -
A(t))\nabla u, \nabla v)_{\Omega}$ .
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By (2.13) there exists a positive constant $\lambda$ such that,
for any sufficiently small $t>0$ ,
$\lambda\Vert\nabla v\Vert_{L^{2}(\Omega)}^{2}\leq(A(t)\nabla v,
\nabla v)_{\Omega}$ , $\forall v\in H_{0}^{1}(\Omega)$ .
Inserting $v=\beta_{t}$ into the above equation, we obtain
$\frac{\lambda}{t^{2}}\Vert\nabla\beta_{t}\Vert_{L^{2}(\Omega)}\leq\Vert
1-A(t)\Vert_{L^{\infty}(\Omega)}(\frac{\Vert\nabla\eta_{t}||_{L^{2}(\Omega)}}{t^{2}}+\Vert\nabla..c\Vert_{L^{2}(\Omega)})$
$+ \frac{1}{t}\Vert
1+t\mathcal{A}^{f}-A(t)\Vert_{L^{\infty}(\Omega)}\Vert\nabla\dot{u}_{\mathcal{L}}\Vert_{L^{2}(\Omega)}+\frac{1}{t^{2}}\Vert\alpha_{t}\Vert_{L^{\infty}(\Omega)}\Vert\nabla
u\Vert_{L^{2}(\Omega)}$ .
Therefore, we conclude $\lim_{tarrow
0+}\Vert\nabla\beta_{t}\Vert_{L^{2}(\Omega)}/t^{2}=0$ and complete
the proof since $\chi_{t}=\beta_{t}+\eta_{t}$ . $\square$
2.3 The Relationship between Eulerian and Lagrangian
derivatives
In this subsection we consider the relationship between Eulerian
and Lagrangian derivatives.From (2.4) we immediately notice
$\dot{u}_{\mathcal{L}}=\dot{u}_{\mathcal{E}}+S\cdot\nabla u$ in
$\Omega$ .
Since $(V\dot{u}_{\mathcal{L}}, \nabla v)=-(\mathcal{A}’\nabla
u, \nabla v)$ and $(V\dot{u}\mathcal{E}, \nabla v)=0$ for any $v\in
H_{0}^{1}(\Omega)$ , we have
$(\nabla (S. \nabla u), \nabla v)_{\Omega}=-(\mathcal{A}’\nabla
u, \nabla v)_{\Omega}$ , $\forall v\in H_{0}^{1}(\Omega)$ .
Similarly, from (2.5) we obtain
$\ddot{u}_{\mathcal{L}}=\ddot{u}_{\mathcal{E}}+2S\cdot\nabla\dot{u}_{\mathcal{E}}+T\cdot\nabla
u+\mathcal{H}_{x}u$ . (S)2 in $\Omega$ .
Since$(\nabla\ddot{u}_{\mathcal{L}}, \nabla
v)_{\Omega}=-(2\mathcal{A}’\nabla\dot{u}_{\mathcal{L}}+\mathcal{A}’’\nabla
u, \nabla v)_{\Omega}$, $(\nabla\ddot{u}_{\mathcal{E}}, \nabla
v)=0$ , $\forall v\in H_{0}^{1}(\Omega)$ ,
we have, for any $v\in H_{0}^{1}(\Omega)$ ,
$(2\nabla(S\cdot Vu_{\mathcal{E}})+\nabla(T\cdot\nabla
u)+\nabla(\mathcal{H}_{x}u\cdot(S)^{2}), \nabla
v)_{\Omega}=-(2\mathcal{A}’\nabla\dot{u}_{\mathcal{L}}+\mathcal{A}’’\nabla
u, \nabla v)_{\Omega}$ .
2.4 Liouville’s Theorem
In this section we prepare Liouville‘s theorem which plays an
important role in calculus ofHadamard’s variation. Following
Garabedian [1] and (2.2), we denote normal components of $S$and $T$
by $\delta\rho$ and $\delta^{2}\rho$ , respectively:
(2.14) $\delta\rho$ $:=S\cdot\nu$ , $\delta^{2}\rho$
$:=T\cdot\nu=\nu^{t}DS(x)S(x)$ .
Theorem 2.2 (Liouville’s Theorem) Let a sufficiently $srn$ooth
function $c(x, t)$ be defined onthe domain $\Omega_{t}$
$:=\mathcal{T}_{t}(\Omega)$ for each $t\geq 0$ . Suppose also that
$c(x, t),$ $c_{t}(x, t)$ $:= \frac{\partial c}{\partial t}(x, t)$
aremeasurable on $\Omega_{t}$ . Then, the following holds:
(2.15) $\frac{d}{dt}(\int_{\Omega_{f}}c(x,
t)dx)|_{t=0}=\int_{\Omega}(c_{t}(x, 0)+\nabla\cdot(c(x,
0)S(x)))dx$
$= \int_{\Omega}c_{t}(x,0)dx+\langle c(\cdot,0),$
$\delta\rho\rangle_{\partial\Omega}$ .
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Proof. We may suppose without loss of generality that
$\partial\Omega,$ $c,$ $c_{t}$ are all sufficiently smooth.The
proof for general cases follows from the density property of
$C^{\infty}(\tilde{\Omega})$ in $H^{1}(\tilde{\Omega})$ . Let
$J\mathcal{T}_{t}(x)$be the Jacobi matrix of $\mathcal{T}_{t}(x)$ .
Differentiating the both sides of
(2.16) $\int_{\Omega_{f}}c(x,
t)dx=\int_{\Omega}c(\mathcal{T}_{t}(x),
t)\det(J\mathcal{T}_{t}(x))dx$
with respect to $t$ , we have
$\frac{d}{dt}(\int_{\Omega_{t}}c(x,
t)dx)=\int_{\Omega}(c_{t}(\mathcal{T}_{t}(x), t)+\nabla
c(\mathcal{T}_{t}(x), t)\cdot\frac{\partial}{\partial
t}\mathcal{T}_{t}(x))\det(/\mathcal{T}_{t}(x))dx$
(2.17)$+
\int_{\Omega}c(\mathcal{T}_{t}(x).t)\frac{\partial}{\partial
t}\det(/\mathcal{T}_{t}(x))dx$ .
Then, letting $tarrow 0+$ , we obtain the first equality of
(2.15). Here, we use
(2.18) $\frac{\partial}{\partial
t}\det(/\mathcal{T}_{t}(x))|_{t=0}=\nabla\cdot S(x)$ .
The second equality immediately follows from the divergence
theorem. $\square$
Corollary 2.3 Suppose that a function $f(x, t)$ is in
$H^{1}(\Omega_{t})$ for each $t\geq 0$ and harmonic on
$\Omega_{t}$with respect to $x\in \mathbb{R}^{n}$ . Then, we
have
$\frac{d}{dt}(\int_{\Omega_{t}}|\nabla_{x}f(x,
t)|^{2}dx)t=0^{=2}\langle\frac{\partial
f}{\partial\nu},\dot{f}_{\mathcal{E}}\rangle_{\partial\Omega}+\langle|\nabla
f|^{2},$$\delta\rho\rangle_{\partial\Omega}$ .
where
$f(x):=f(x,0),\dot{f}_{\mathcal{E}}(x):=\partial\overline{t}\partial
f(x, t)|_{t=0}$.
Proof: Set $c(x, t);=|\nabla_{x}f(x, t)|^{2}$ and apply Theorem
2.2. $\square$
We now try to obtain a second order Liouville’s theorem. Assume
that $\partial\Omega,$ $c,$ $S$ are suffi-ciently smooth. We have
obtained (2.17) by differentiating the both side of (2.16) with
respectto $t$ . One more differentiation of the both side of (2.17)
and letting $tarrow 0$ yield
$\int_{\Omega}[c_{tt}(x, 0)+2\nabla_{x}c_{t}(x, 0)\cdot
S+\nabla_{x}c(x, 0)\cdot T+\mathcal{H}_{x}c(x, 0) . (S)^{2}]dx$
$+2 \int_{\Omega}[c_{t}(x,0)+\nabla_{x}c(x, 0)\cdot
S](\nabla\cdot S)dx$
$+ \int_{\Omega}c(x, 0)[\nabla\cdot T+2\sum_{i
-
We try to simplify this formula. Recall that $DS$ is the Jacobi
matrix of the vector field S. Sinceit follows from the divergence
theorem that
2 $\int_{\Omega}(\nabla_{x}c(x, 0)\cdot S)(\nabla\cdot
S)dx=2\int_{\partial\Omega}(\nabla_{x}c(x,0)\cdot S)\delta\rho
ds$
$-2
\int_{\Omega}\mathcal{H}_{x}c(x,0)\cdot(S)^{2}dx-2\int_{\Omega}((DS)S)\cdot\nabla_{x}c(x,0)dx$
,
we have
$\frac{d^{2}}{dt^{2}}(\int_{\Omega},$ $c(x, t)
dx)t=0=\int_{\Omega}c_{tt}(x, 0)dx+2\int_{\partial\Omega}c_{t}(x,
0)\delta\rho ds$
$+ \int_{\partial\Omega}c(x,0)\delta^{2}\rho
ds+2\int_{\partial\Omega}(\nabla_{x}c(x, 0)\cdot S)\delta\rho
ds$
$-
\int_{\Omega}\mathcal{H}_{x}c(x,0)\cdot(S)^{2}dx-2\int_{\Omega}((DS)S)\cdot\nabla_{x}c(x,
0)dx$
$+2 \int_{\Omega}c(x,0)\sum_{i
-
Therefore, we have
$X+Y+Z=- \int_{\partial\Omega}(\nabla_{x}c\cdot
S)\delta\rho+\sum_{i
-
Here, $\omega_{n}$ is the measure of $(n-1)$-dimensional sphere
$S^{n-1}$ . Then, for sufficiently smoothfunction $f$ we have
Green’s formula
(3.1) $- \int_{\Omega}\Gamma(x-y)\Delta
f(x)dx+\int_{\partial\Omega}\frac{\partial
f}{\partial\nu}(x)\Gamma(x-y)ds_{x}=\int_{\partial\Omega}f(x)\frac{\partial}{\partial\nu_{x}}\Gamma(x-y)ds_{x}+f(y)$.
For the fundamental solution $\Gamma(x-y)$ . define $u$ as a
solution of the following boundary valueproblem:
$\Delta u=0$ in $\Omega$ , $u(x)=-\Gamma(x-y)$ ,
$x\in\partial\Omega$ .Then,
$G(x, y):=\Gamma(x-y)+u(x)$
is the Green function of $\triangle$ on $\Omega$ . It follows
from the definition that $G(x,y)=0$ for $x\in\partial\Omega$
and$y\in\Omega$ . Adding the following Green’s formula with respect
to $f$ and $u$
$- \int_{\Omega}u(x)\Delta
f(x)dx+\int_{\partial\Omega}\frac{\partial
f}{\partial\nu}(x)u(x)ds=\int_{\partial\Omega}f(x)\frac{\partial
u}{\partial\nu}(x)ds$
to (3.1), we obtain Green $s$ second formula
(3.2) $f(y)=- \int_{\Omega}G(x,y)\Delta
f(x)dx-\int_{\partial\Omega}f(x)\frac{\partial}{\partial\nu_{x}}G(x,
y)ds_{x}$.
3.1 First Variation
Now, we consider domain perturbation
$\Omega_{t}=\mathcal{T}_{t}(\Omega)$ defined in the previous
section. The Greenfunction $G(x, y, t)$ on $\Omega_{t}$ is written
as
$G(x, y, t)=\Gamma(x-y)+u(x,t)$ ,
where $u(x, t)$ is the harmonic function which satisfies
(3.3) $\triangle_{x}u(x, t)=0$ in $\Omega_{t}$ , $u(x,
t)=-\Gamma(x-y)$ , $x\in\partial\Omega_{t}$ .
Obviously, we have $G(x, y, 0)=G(x, y)$ and $u(x,0)=u(x)$ . For
two inner points $x,$ $y\in\Omega$ andsufficiently small $t>0$ ,
we have $x,$ $y\in\Omega_{t}$ . The first variation $\delta G(x,
y)$ with respect to domainperturbation is defined by
$\delta G(x, y):=\lim_{tarrow
0+}\frac{G(x,y,t)-G(x,y,0)}{t}=\lim_{tarrow
0+}\frac{u(x,t)-u(x,0)}{t}=\dot{u}_{\mathcal{E}}(x)$ ,
and is equal to the Eulerian derivative $\dot{u}\mathcal{E}$ of
$u$ .By (2.6), we confirm that $\dot{u}\mathcal{E}$ is a solution
of the boundary value problem
$\triangle\dot{u}\mathcal{E}=0$ in $\Omega$ ,
$\dot{u}\mathcal{E}=S\cdot(-\nabla_{x}\Gamma(x-y)-\nabla
u)=-S\cdot\nabla_{x}G(x,y)=-\delta\rho\frac{\partial}{\partial\nu_{x}}G(x,y)$
on $\partial\Omega$ .
Here, we use the fact that $S\cdot\nabla_{x}G(x,
y)=(S\cdot\nu)\frac{\partial}{\partial\nu_{x}}G(x, y)$ on
$\partial\Omega$ . Applying the formula (3.2)to
$\dot{u}_{\mathcal{E}}$ , we obtain Hadamard $s$ variational
formula.
Theorem 3.1 (Hadamard’s variational formula) The first
variation. $\delta G(w, y)$ of the Greenfunction $G(w, y)$ of
$\triangle$ with respect to domain perturbation is given by
$\delta
G(w,y)=\int_{\partial\Omega}\frac{\partial}{\partial\nu_{x}}G(x,
y)\frac{\partial}{\partial\nu_{x}}G(x, w)\delta\rho ds_{x}$ ,
$\delta\rho:=S\cdot\nu$ .
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3.2 Second VariationIn this subsection, we compute the second
variation of the Green function with respect to domainperturbation.
We prepare a lemma. Let a harmonic function $u(x, t)$ be a solution
of the Dirichletproblem (3.3). Since $\delta G(x,
y)=\dot{u}_{\mathcal{E}}(x)$ , we recall that
(3.4) $\delta G(x,
y)=-\delta\rho\frac{\partial}{\partial\nu_{x}}G(x, y)$ ,
$x\in\partial\Omega$ .
Hence, for a harmonic function $g(x)$ , we find
$0= \int_{\partial\Omega}(\delta G(x,
y)+\delta\rho\frac{\partial}{\partial\nu_{x}}G(x, y))\frac{\partial
g}{\partial\nu}(x)ds_{x}$
$= \int_{\Omega}\nabla_{x}\delta G(x, y)\cdot\nabla
g(x)dx+\int_{\partial\Omega}\delta\rho\frac{\partial}{\partial\nu_{x}}G(x,
y)\frac{\partial g}{\partial\nu}(x)ds_{x}$ ,
and obtain the following lemma.
Lemma 3.2 For a harmonic function $g$ on $\Omega$ , the
following equality holds:
$\int_{\Omega}\nabla_{x}\delta G(x, y)\cdot\nabla
g(x)dx=-\int_{\partial\Omega}\frac{\partial}{\partial\nu_{x}}G(x,
y)\frac{\partial g}{\partial\nu}(x)\delta\rho ds_{x}$ .
The second variation $\delta^{2}G(x, y)$ of the Green function
$G(x, y)$ is defined by
$\delta^{2}G(x, y):=\frac{\partial^{2}}{\partial t^{2}}G(x, y,
t)|_{t=0}=\ddot{u}_{\mathcal{E}}(x)$ ,
and, therefore, we only need to compute $\ddot{u}_{\mathcal{E}}$
. Recall that the harmonic function is a solution ofthe Dirichlet
problem
$\triangle u=0$ in $\Omega$ , $u=-\Gamma(\cdot-y)$ on
$\partial\Omega$ .
By (2.7), the boundary value of $\ddot{u}_{\mathcal{E}}$ on
$\partial\Omega$ is
$\ddot{u}\mathcal{E}=-2S\cdot\nabla_{x}\delta G(x,
y)-T\cdot\nabla_{x}G(x, y)-\mathcal{H}_{x}G(x, y)\cdot(S)^{2}$
.
Here, we use $G(x, y)=\Gamma(x-y)+u(x)$ and
$\dot{u}_{\mathcal{E}}(x)=\delta G(x, y)$ . From (3.2), we find
$\delta^{2}G(x,
y)=\ddot{u}s(x)=2\int_{\partial\Omega}S\cdot\nabla_{w}\delta G(w,
y)\frac{\partial}{\partial\nu_{x}}G(x, w)ds_{w}$
(3.5) $+ \int_{\partial\Omega}T\cdot\nabla_{w}G(w,
y)\frac{\partial}{\partial\nu_{x}}G(x, w)ds_{w}$
$+ \int_{\partial\Omega}\mathcal{H}_{w}G(w,
y)\cdot(S)^{2}\frac{\partial}{\partial\nu_{x}}G(x, w)ds_{w}$ .
We denote the first, second and third terms of the right-hand
side of (3.5) by $X,$ $Y,$ $Z$ , respec-tively. As before, the term
$Y$ can be written as
(3.6) $Y=
\int_{\partial\Omega}\frac{\partial}{\partial\nu_{w}}G(w,
y)\frac{\partial}{\partial\nu_{x}}G(x, w)\delta^{2}\rho ds_{w}$
.
To understand the terms $X$ and $Z$ , we consider the
$(n-1)$-dimensional tangent space $\prime 1_{x}’\partial\Omega$of
$\partial\Omega$ at $x\in\partial\Omega$ . Let $\{s_{1}, \cdots ,
s_{n-1}\}$ be the orthonormal basis of $\prime 1_{x}\partial\Omega$
. Then, $\{s_{1}, \cdots, s_{n-1}, \nu\}$
137
-
is an orthonormal basis of the tangent space $\prime
1_{x}\mathbb{R}^{n}$ at $x\in \mathbb{R}^{n}$ . For a generic
function $f$ , directionalderivatives are defined by
$\frac{\partial f}{\partial\nu}=\nabla f\cdot\nu$ ,
$\frac{\partial f}{\partial s_{i}}=\nabla f\cdot s_{i}$ , $i=1,$
$\cdots,$ $n-1$ .
Thus, defining the orthogonal matrix $P$ by $P:=(s_{1}\ldots.,
s_{n-1}, \nu)$ , we may write
$( \frac{\partial f}{\partial s_{1}},$ $\cdot\cdot\cdot$
$\frac{\partial f}{\partial s_{n-1}},$ $\frac{\partial
f}{\partial\nu})=(\nabla f)P$, or
(3.7) $( \nabla f)^{T}=\sum_{i=1}^{n-1}s_{i}\frac{\partial
f}{\partial s_{i}}+\nu\frac{\partial f}{\partial\nu}$ and
$\nabla=\sum_{i=1}^{n-1}s_{i}^{T}\frac{\partial}{\partial
s_{i}}+\nu^{T}\frac{\partial}{\partial\nu}$.
If we write $S$ as
(3.8) $S=\sum_{i=1}^{n-1}\mu_{i}s_{i}+\delta\rho\nu$ ,
$\delta\rho=S\cdot\nu$ , $\mu_{i}=S$ . si, $i=1,$ $\cdots,$
$7t-1$
on $\partial\Omega$ , we obtain
$S\cdot\nabla_{w}\delta G(w,
y)=\sum_{i=1}^{n-1}\mu_{i}\frac{\partial}{\partial s_{i}}\delta
G(w, y)+\delta\rho\frac{\partial}{\partial\nu_{w}}\delta G(w,
y)$.
Using Lemma 3.2 with $g:=\delta G$ , the term $X$ (the first
term of the right-hand side of (3.5)) iswritten as
$X=2
\sum_{i=1}^{n-1}\int_{\partial\Omega}\mu_{i}\frac{\partial}{\partial
s_{i}}\delta G(w,y)\frac{\partial}{\partial\nu_{x}}G(x,
w)ds_{w}+2\int_{\partial\Omega}\delta\rho\frac{\partial}{\partial\nu_{w}}\delta
G(w,y)\frac{\partial}{\partial\nu_{x}}G(x, w)ds_{w}$
$=2
\sum_{i=1}^{n-1}\int_{\partial\Omega}\mu_{i}\frac{\partial}{\partial
s_{i}}\delta G(w,y)\frac{\partial}{\partial\nu_{x}}G(x,
w)ds_{w}-2\int_{\Omega}\nabla_{w}\delta G(w,
y)\cdot\nabla_{x}\delta G(x, w)ds_{w}$.
Next, we try to rewrite the third term $Z$ of the right-hand
side of (3.5). To this end, weconsider the curved coordinate
defined by $\{s_{1}, \cdots, s_{n-1}, \nu\}$ in the neighborhood of
$x\in\partial\Omega$ andsecond order differentiation on the
coordinate. For a $C^{2}$ class generic function $f$ , the
Hessematrix $\mathcal{H}f$ is written by
$\mathcal{H}f=\nabla(\nabla f)^{T}$ and
$\mathcal{H}f=\nabla(\sum_{i=1}^{n-1}s_{i}\frac{\partial
f}{\partial s_{i}}+\nu\frac{\partial f}{\partial\nu})$
$= \sum_{i=1}^{n-1}\frac{\partial f}{\partial
s_{i}}Ds_{i}+\frac{\partial
f}{\partial\nu}D\nu+\sum_{i=1}^{n-1}s_{i}\nabla(\frac{\partial
f}{\partial s_{i}})+\nu\nabla(\frac{\partial f}{\partial\nu}I$
(3.9) $= \sum_{i=1}^{n-1}\frac{\partial f}{\partial
s_{i}}Ds_{i}+\frac{\partial
f}{\partial\nu}D\nu+\sum_{i,j=1}^{n-1}\frac{\partial^{2}f}{\partial
s_{j}\partial
s_{i}}s_{i}s_{j}^{T}+\sum_{i=1}^{n-1}\frac{\partial^{2}f}{\partial
s_{i}\partial\nu}(s_{i}\nu^{T}+\nu
s_{i}^{T})+\frac{\partial^{2}f}{\partial\nu^{2}}\nu\nu^{T}$.
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-
Similarly, $\triangle f$ is $co\dot{r}nputed$ as
$\triangle f=\nabla\cdot\nabla
f=\nabla\cdot(\sum_{i=1}^{n-1}s_{i}\frac{\partial f}{\partial
s_{i}}+\nu\frac{\partial f}{\partial\nu})$
$=( \nabla\cdot\nu)\frac{\partial
f}{\partial\nu}+\nu\cdot\nabla\frac{\partial
f}{\partial\nu}+\sum_{i=1}^{n-1}((\nabla\cdot s_{i})\frac{\partial
f}{\partial s_{i}}+s_{i}\cdot\nabla\frac{\partial f}{\partial
s_{i}})$
(3.10) $=( \nabla\cdot\nu)\frac{\partial
f}{\partial\nu}+\frac{\partial^{2}f}{\partial\nu^{2}}+\sum_{i=1}^{n-1}((\nabla\cdot
s_{i})\frac{\partial f}{\partial
s_{i}}+\frac{\partial^{2}f}{\partial s_{i}^{2}})$ .
Since the Green function $G$ satisfies $G(x, y)=0$ on
$\partial\Omega$ , we have
$\frac{\partial}{\partial s_{i}}G(x,
y)=\frac{\partial^{2}}{\partial s_{i}\partial s_{j}}G(x, y)=0$,
$i,j=1,$ $\cdots,$ $?-1$ .
and, thus,
$0= \triangle_{w}G(w,
y)=(\nabla\cdot\nu)\frac{\partial}{\partial\nu_{w}}G(w,
y)+\frac{\partial^{2}}{\partial\nu_{w}^{2}}G(w, y)$ ,
$w\in\partial\Omega$ .
Applying these results to computation of $\mathcal{H}_{w}G(w,
y)\cdot S^{2}$ with $S=\sum_{i=1}^{n-1}\mu_{i}s_{i}+\delta\rho\nu$
, we obtain
$\mathcal{H}_{w}G(w,
y)\cdot(S)^{2}=2\delta\rho\sum_{i=1}^{n-1}\mu_{i}\frac{\partial^{2}}{\partial
s_{i}\partial\nu_{w}}G(w,y)-(\nabla\cdot\nu)(\delta\rho)^{2}\frac{\partial}{\partial\nu_{w}}G(w,
y)$
$+(
\sum_{i,j=1}^{n-1}\mu_{i}\mu_{j}s_{i}^{T}(D\nu)s_{j}+2\sum_{i=1}^{n-1}\mu_{i}\delta
ps_{i}^{T}(D\nu)\nu+(\delta\rho)^{2}\nu^{T}(D\nu)\nu)\frac{\partial}{\partial\nu_{w}}G(w,
y)$
$=2
\delta\rho\sum_{i=1}^{n-1}\mu_{i}\frac{\partial^{2}}{\partial
s_{i}\partial\nu_{w}}G(w,
y)+(\sum_{i=1}^{n-1}\kappa_{i}(\mu_{i}^{2}-(\delta\rho)^{2}))\frac{\partial}{\partial\nu_{w}}G(w,
y)$.
Here, we use the fact
$\nabla\nu=\sum_{i=1}^{n-1}\kappa_{i}s_{i}s_{i}^{t},$
$\nabla\cdot\nu=$ tr $( \nabla\nu)=\sum_{i=1}^{n-1}\kappa_{i}$ ,
where $\kappa_{i}$ is the curvatureof the cross-section of
$(n-1)$-dimensional surface $\partial\Omega$ by a two-dimensional
plane defined by $s_{i}$and $\nu$ . Therefore, the third term $Z$
of the right-hand side of (3.5) is written by
$Z=
\int_{\partial\Omega}\sum_{i=1}^{n-1}\kappa_{i}(\mu_{i}^{2}-(\delta\rho)^{2})\frac{\partial}{\partial\nu_{w}}G(w,
y)\frac{\partial}{\partial\nu_{x}}G(x, w)ds_{w}$
$+2
\int_{\partial\Omega}\sum_{i=1}^{n-1}\mu_{i}\delta\rho\frac{\partial^{2}}{\partial
s_{i}\partial\nu_{w}}G(w.y)\frac{\partial}{\partial\nu_{x}}G(x,
w)ds_{w}$ .
Noticing (3.4), we see the equality
$\frac{\partial}{\partial s_{i}}\delta G(w,
y)+\delta\rho\frac{\partial^{2}}{\partial s_{i}\partial\nu_{w}}G(w,
y)=-\frac{\partial(\delta\rho)}{\partial
s_{i}}\frac{\partial}{\partial\nu_{w}}G(w, y)$.
Computing $X+Z$ using this equality, we find
$X+Z=-2 \int_{\Omega}\nabla_{w}\delta G(w, y)\nabla_{w}\delta
G(x, w)ds_{w}$
(3.11)$+
\int_{\partial\Omega}\sum_{i=1}^{n-1}[\kappa_{i}(\mu_{i}^{2}-(\delta\rho)^{2})-2\mu_{i}\frac{\partial(\delta\rho)}{\partial
s_{i}}]\frac{\partial}{\partial\nu_{w}}G(w,
y)\frac{\partial}{\partial\nu_{x}}G(x.e\iota)ds_{w}$ .
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-
Gathering (3.5), (3.6), and (3.11) we obtain
$\delta^{2}G(x, y)=-2\int_{\Omega}\nabla_{w}\delta G(w,
y)\nabla_{w}\delta G(x,
w)ds_{w}+\int_{\partial\Omega}\chi\frac{\partial}{\partial\nu_{w}}G(w,
y)\frac{\partial}{\partial\nu_{x}}G(x, w)ds_{w}$,
$\chi:=\delta^{2}\rho+\sum_{i=1}^{n-1}[\kappa_{i}(\mu_{i}^{2}-(\delta\rho)^{2})-2\mu_{i}\frac{\partial(\delta\rho)}{\partial
s_{i}}]$ .
We further try to simplify $\chi$ . At first, since
$\partial\partial t=\kappa_{i}s_{i}$ , we notice
$\sum_{i=1}^{n-1}\kappa_{i}\mu_{i}^{2}=\sum_{i=1}^{n-1}\kappa_{i}(S\cdot
s_{i})^{2}=\sum_{i=1}^{n-1}S\cdot
s_{i}(\frac{\partial(\delta\rho)}{\partial s_{i}}-$ $(
\frac{\partial S}{\partial s_{i}}$ . $\nu))$ .
Thus, recalling from (3.7) that $D
S=\sum_{n=1\mathcal{T}s_{i}}^{n-1\partial
S}s_{i}^{T}+\frac{\partial S}{\partial\nu}\nu^{T}$ , we see
$\sum_{i=1}^{n-1}S\cdot s_{i}(\frac{\partial S}{\partial
s_{i}}\cdot\nu)=\sum_{i=1}^{n-1}S\cdot
s_{i}(\nu^{T}(DS)s_{i})=\nu^{T}(DS)(\sum_{i=1}^{n-1}(S\cdot
s_{i})s_{i})$
(3.12)
$=
\nu^{T}(DS)(S-\delta\rho\nu)=\delta^{2}\rho-\delta\rho(\frac{\partial
S}{\partial\nu}\cdot\nu)=\delta^{2}\rho-\frac{1}{2}\frac{\partial(\delta^{-}\rho)^{2}}{\partial\nu}$
.
Here, we use the facts $\delta^{2}\rho=\nu^{T}(DS)S$ by (2.14)
and $\frac{\partial\nu}{\partial\nu}=0$ . Similarly, since
$S\cdot\nabla=\sum_{n=1}^{n-1}(S\cdot$$s_{i})_{Ts_{i}}^{\partial}+(S\cdot\nu)_{T\nu}^{\partial}$
by (3.7), we have
(3.13) $\sum_{i=1}^{n-1}S\cdot
s_{i}\frac{\partial(\delta\rho)}{\partial
s_{i}}=(S\cdot\nabla)\delta\rho-\delta\rho\frac{\partial(\delta\rho)}{\partial\nu}=(S\cdot\nabla)\delta\rho-\frac{1}{2}\frac{\partial(\delta\rho)^{2}}{\partial\nu}$
.
Letting $\tilde{\kappa}$ $:= \sum_{i=1}^{n-1}\kappa_{i},$ $\chi$
is rewritten as
$\chi=-\tilde{\kappa}(\delta\rho)^{2}-(S\cdot\nabla)\delta
p+\frac{\partial(\delta^{-}\rho)^{2}}{\partial\nu}$ .
If the domain perturbation $\mathcal{T}_{t}(x)$ satisfies
$S\cdot s_{i}=0,$ $i=1,$ $\cdots,$ $n-1$ , it follows from
(3.12),(3.13) that
(3.14)
$\delta^{2}\rho=(S\cdot\nabla)\delta\rho=\frac{1}{2}\frac{\partial(\delta\rho)^{2}}{\partial\nu}$
.
Therefore, in this case, we
find$\chi=-\tilde{\kappa}(\delta\rho)^{2}+\delta^{2}\rho$.
So far, computation has been done under smoothness assumptions.
A usual density argu-ment yields the following theorem:
Theorem 3.3 (Hadamard’s second variational formula) Let
$\Omega\subset \mathbb{R}^{n}$ be a Lipschitz do-main and
$\mathcal{T}_{t}(x)$ be a $W^{1,\infty}$ class domain perturbation
on $\Omega$ . Then, the second variation $\delta^{2}G(w, y)$of the
Green function $G(w, y)$ of Laplacian on $\Omega$ is written by
$\delta^{2}G(x, y)=\langle\chi\frac{\partial}{\partial\nu}G(x,
\cdot),$ $\frac{\partial}{\partial\nu}G(\cdot,
y)\rangle_{\partial\Omega}-2(\nabla\delta G(x, \cdot), \nabla\delta
G(\cdot, y))_{\Omega}$ ,
$\chi=-\tilde{\kappa}(\delta
p)^{2}-(S\cdot\nabla)\delta\rho+\frac{\partial(\grave{\delta}\rho)^{2}}{\partial\nu}$
, $\delta\rho:=S\cdot\nu$ ,
$\tilde{\kappa}:=\sum_{i=1}^{n-1}t\mathfrak{i}_{i}$ ,
140
-
where $\{s_{i}\}_{i=1}^{n-1}$ is an orthonormal basis of the
tangent space of $\partial\Omega$ and $\kappa_{i}$ is the curvature
of $\partial\Omega$along si. In particular, if the perturbation
satisfies S. $s_{i}=0,$ $i=1,$ $\cdots,$ $n-1$ , we have
(3.15) $\delta^{2}G(x,
y)=\langle(\delta^{2}\rho-\tilde{\kappa}(\delta\rho)^{2})\frac{\partial}{\partial\nu}G(x,
\cdot),$ $\frac{\partial}{\partial\nu}G(\cdot,
y)\rangle_{\partial\Omega}-2(\nabla\delta G(x, \cdot), \nabla\delta
G(\cdot, y))_{\Omega}$ .
Remark: Garabedian and Schiffer [2] dealt with domain
perturbation such as
$\mathcal{T}_{t}(x)=x+th(x)\nu(x)$ , $t\geq 0$ ,
where $h(x)$ is a scalar function defined on $\partial\Omega$ .
In this case, Hadamard’s second variationalformula is (3.15) with
$\delta\rho=h$ and $\delta^{2}\rho(x)=0$ which is exactly same to
Garabedian-Schiffer’sformula [2]. Therefore, Theorem 3.3 is an
extension of Garabedian-Schiffer’s formula.
References
[1] P.R. GARABEDIAN, Partial Differential Equations (2nd ed.)
(1986) Chelsea.
[2] P.R. GARABEDIAN, M. SCHIFFER, Convexity of domain
functionals, J. Anal. Math., 2(1952-53), 281-368.
[3] J. HADAMARD, M\’emoire sur le probl\‘eme $d$ ‘analyse
relatif \‘a 1‘\’equilibre des plaques \’elastiquesencastrees.
M\’emoires present\’es par divers savants \‘a 1‘Acad\’emie des
Sciences, Vol. 33 (1908),1-128 (Oeuvres., 2 (1968), 515-631).
[4] M. SCHIFFER, Hadamard $s$ formula and variation of
domain-functions, Amer. J. Math., 68(1946), 417-448.
141